answer:Given that (triangle ABC) is a triangle with circumcenter (O), from vertex (A) draw the altitude to (BC) with foot (P). It is also provided that (angle BCA geq angle ABC + 30^circ). We need to prove that (angle CAB + angle COP < 90^circ). 1. Define the angles as follows: [ alpha = angle CAB, quad beta = angle ABC, quad gamma = angle BCA, quad delta = angle COP. ] 2. From the definition of the angles, we need to show that: [ alpha + delta < 90^circ. ] 3. Observing the circumcircle (triangle ABC), we recognize: [ alpha = frac{1}{2} angle BOC ] 4. From the law of angles in a circle, [ angle BOC = 180^circ - 2 angle BCA ] 5. Since (alpha = 1/2 angle BOC), we substitute: [ alpha = frac{1}{2} (180^circ - 2 gamma) = 90^circ - gamma ] 6. Substitute (gamma = angle PCO): [ alpha = 90^circ - gamma ] 7. Hence, it delineates the focus of our proof to show: [ gamma > delta ] 8. Keeping with the properties of the triangle, let's introduce points. Define (K) and (Q) as the symmetric points of (A) and (P) respectively with respect to the perpendicular bisector of (BC). Let (R) be the circumradius of (triangle ABC). Now, we know: [ OA = OB = OC = OK = R ] 9. Because (KQPA) forms a rectangle, we know: [ QP = KA ] 10. Considering the angles in (triangle text{OAB}), we derive: [ angle AOK = angle AOB - angle KOB = angle AOB - angle AOC = 2gamma - 2beta geq 60^circ ] 11. Since (angle AOK geq 60^circ) and (OA = OK = R), it follows: [ KA geq R, quad QP geq R ] 12. Utilizing the triangle inequality in (triangle OOQ) entails: [ OP + R = OQ + OC > QC = QP + PC geq R + PC ] 13. Therefore: [ OP > PC ] 14. Within (triangle COO), it is inferred: [ angle PCO > delta ] 15. Combining the derived angles: [ alpha = 90^circ - angle PCO ] 16. Hence: [ angle CAB + angle COP = alpha + delta = 90^circ - gamma + gamma < 90^circ ] # Conclusion: (boxed{<90^circ})

answer:Let's break down the journey into two parts: 1. First part of the journey: Speed = 40 kmph Time = 3 hours We can calculate the distance covered in the first part using the formula: Distance = Speed × Time So, the distance covered in the first part is: Distance = 40 kmph × 3 hours = 120 km 2. Second part of the journey: We know the total distance of the journey is 240 km. Since we've already covered 120 km in the first part, the remaining distance for the second part is: Total distance - Distance covered in the first part = 240 km - 120 km = 120 km The total time for the journey is 5 hours, and the first part took 3 hours, so the time taken for the second part is: Total time - Time taken for the first part = 5 hours - 3 hours = 2 hours Now, let's find the speed during the second part of the journey using the formula: Speed = Distance / Time The speed during the second part is: Speed = 120 km / 2 hours = 60 kmph So, the speed during the second part of the journey was boxed{60} kmph.

answer:Since 2^{0.5}>2^{0}=1, 0<log_{pi}3<log_{pi}pi=1, and log_{2}0.5<log_{2}1=0, Therefore, a>b>c. Hence, the correct option is boxed{A}. This can be determined by using the properties of exponential and logarithmic functions. Mastering the properties of exponential and logarithmic functions is key to solving the problem.

answer:For ①, its converse proposition is: If x>|y|, then x>y, which is a true proposition. This is because if x>|y|, it implies x>y since |y| is always greater than or equal to y. For ②, its negation is: If xleq1, then x^2leq1, which is a false proposition. For example, if x=-5, then x^2=25>1. For ③, its negation is: If xneq1, then x^2+x-2neq0. However, when x=-2, x^2+x-2=0, making it a false proposition. For ④, if x^2>0, then x>0 or x<0, which does not necessarily mean x>1. Therefore, both the original proposition and its contrapositive are false propositions. Therefore, the true proposition among the given options is boxed{1}.